How to solve decimal logarithms and fractions. Definition in mathematics
We continue to study logarithms. In this article we will talk about calculating logarithms, this process is called logarithm. First we will understand the calculation of logarithms by definition. Next, let's look at how the values of logarithms are found using their properties. After this, we will focus on calculating logarithms through the initially specified values of other logarithms. Finally, let's learn how to use logarithm tables. The entire theory is provided with examples with detailed solutions.
Page navigation.
Calculating logarithms by definition
In the simplest cases it is possible to perform quite quickly and easily finding the logarithm by definition. Let's take a closer look at how this process happens.
Its essence is to represent the number b in the form a c, from which, by the definition of a logarithm, the number c is the value of the logarithm. That is, by definition, the following chain of equalities corresponds to finding the logarithm: log a b=log a a c =c.
So, calculating a logarithm by definition comes down to finding a number c such that a c = b, and the number c itself is the desired value of the logarithm.
Taking into account the information in the previous paragraphs, when the number under the logarithm sign is given by a certain power of the logarithm base, you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show solutions to examples.
Example.
Find log 2 2 −3, and also calculate the natural logarithm of the number e 5,3.
Solution.
The definition of the logarithm allows us to immediately say that log 2 2 −3 =−3. Indeed, the number under the logarithm sign is equal to base 2 to the −3 power.
Similarly, we find the second logarithm: lne 5.3 =5.3.
Answer:
log 2 2 −3 =−3 and lne 5,3 =5,3.
If the number b under the logarithm sign is not specified as a power of the base of the logarithm, then you need to carefully look to see if it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the logarithm sign is equal to the base to the power of 1, or 2, or 3, ...
Example.
Calculate the logarithms log 5 25 , and .
Solution.
It is easy to see that 25=5 2, this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2.
Let's move on to calculating the second logarithm. The number can be represented as a power of 7: 
(see if necessary). Hence, 
.
Let's rewrite the third logarithm in the following form. Now you can see that 
, from which we conclude that 
. Therefore, by the definition of logarithm 
.
Briefly, the solution could be written as follows: .
Answer:
log 5 25=2 , 
And .
When there is a sufficiently large natural number under the logarithm sign, it does not hurt to expand it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore calculate this logarithm by definition.
Example.
Find the value of the logarithm.
Solution.
Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of one and the property of the logarithm of a number equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1. That is, when under the sign of the logarithm there is a number 1 or a number a equal to the base of the logarithm, then in these cases the logarithms are equal to 0 and 1, respectively.
Example.
What are logarithms and log10 equal to?
Solution.
Since , then from the definition of logarithm it follows 
.
In the second example, the number 10 under the logarithm sign coincides with its base, so the decimal logarithm of ten is equal to one, that is, lg10=lg10 1 =1.
Answer:
AND lg10=1 .
Note that the calculation of logarithms by definition (which we discussed in the previous paragraph) implies the use of the equality log a a p =p, which is one of the properties of logarithms.
In practice, when a number under the logarithm sign and the base of the logarithm are easily represented as a power of a certain number, it is very convenient to use the formula 
, which corresponds to one of the properties of logarithms. Let's look at an example of finding a logarithm that illustrates the use of this formula.
Example.
Calculate the logarithm.
Solution.
Answer:
.
Properties of logarithms not mentioned above are also used in calculations, but we will talk about this in the following paragraphs.
Finding logarithms through other known logarithms
The information in this paragraph continues the topic of using the properties of logarithms when calculating them. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's give an example for clarification. Let's say we know that log 2 3≈1.584963, then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .
In the above example, it was enough for us to use the property of the logarithm of a product. However, much more often it is necessary to use a wider arsenal of properties of logarithms in order to calculate the original logarithm through the given ones.
Example.
Calculate the logarithm of 27 to base 60 if you know that log 60 2=a and log 60 5=b.
Solution.
So we need to find log 60 27 . It is easy to see that 27 = 3 3 , and the original logarithm, due to the property of the logarithm of the power, can be rewritten as 3·log 60 3 .
Now let's see how to express log 60 3 in terms of known logarithms. The property of the logarithm of a number equal to the base allows us to write the equality log 60 60=1. On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2·log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2·log 60 2−log 60 5=1−2·a−b.
Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3·(1−2·a−b)=3−6·a−3·b.
Answer:
log 60 27=3·(1−2·a−b)=3−6·a−3·b.
Separately, it is worth mentioning the meaning of the formula for transition to a new base of the logarithm of the form 
. It allows you to move from logarithms with any base to logarithms with a specific base, the values of which are known or it is possible to find them. Usually, from the original logarithm, using the transition formula, they move to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow their values to be calculated with a certain degree of accuracy. In the next paragraph we will show how this is done.
Logarithm tables and their uses
For approximate calculation of logarithm values can be used logarithm tables. The most commonly used base 2 logarithm table is the table natural logarithms and table decimal logarithms. When working in decimal system For calculus, it is convenient to use a table of logarithms based on base ten. With its help we will learn to find the values of logarithms.
The presented table allows you to find the values of the decimal logarithms of numbers from 1,000 to 9,999 (with three decimal places) with an accuracy of one ten-thousandth. We will analyze the principle of finding the value of a logarithm using a table of decimal logarithms into specific example– it’s clearer that way. Let's find log1.256.
In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). The third digit of the number 1.256 (digit 5) is found in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (digit 6) is found in the first or last line to the right of the double line (this number is circled with a green line). Now we find the numbers in the cells of the table of logarithms at the intersection of the marked row and marked columns (these numbers are highlighted orange). The sum of the marked numbers gives the desired value of the decimal logarithm accurate to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.
Is it possible, using the table above, to find the values of decimal logarithms of numbers that have more than three digits after the decimal point, as well as those that go beyond the range from 1 to 9.999? Yes, you can. Let's show how this is done with an example.
Let's calculate lg102.76332. First you need to write down number in standard form : 102.76332=1.0276332·10 2. After this, the mantissa should be rounded to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm the resulting number, that is, we take log102.76332≈lg1.028·10 2. Now we apply the properties of the logarithm: lg1.028·10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.
In conclusion, it is worth noting that using a table of decimal logarithms you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.
For example, let's calculate log 2 3 . According to the formula for transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, 
.
Bibliography.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).
What is a logarithm?
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
What is a logarithm? How to solve logarithms? These questions confuse many graduates. Traditionally, the topic of logarithms is considered complex, incomprehensible and scary. Especially equations with logarithms.
This is absolutely not true. Absolutely! Don't believe me? Fine. Now, in just 10 - 20 minutes you:
1. You will understand what is a logarithm.
2. Learn to solve a whole class exponential equations. Even if you haven't heard anything about them.
3. Learn to calculate simple logarithms.
Moreover, for this you will only need to know the multiplication table and how to raise a number to a power...
I feel like you have doubts... Well, okay, mark the time! Go!
First, solve this equation in your head:
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Tasks whose solution is converting logarithmic expressions, are quite common on the Unified State Examination.
In order to successfully cope with them with minimal time, in addition to the basic logarithmic identities, you need to know and correctly use some more formulas.
This is: a log a b = b, where a, b > 0, a ≠ 1 (It follows directly from the definition of the logarithm).
log a b = log c b / log c a or log a b = 1/log b a
where a, b, c > 0; a, c ≠ 1.
log a m b n = (m/n) log |a| |b|
where a, b > 0, a ≠ 1, m, n Є R, n ≠ 0.
a log c b = b log c a
where a, b, c > 0 and a, b, c ≠ 1
To show the validity of the fourth equality, let’s take the logarithm of the left and right sides to base a. We get log a (a log with b) = log a (b log with a) or log with b = log with a · log a b; log c b = log c a · (log c b / log c a); log with b = log with b.
We have proven the equality of logarithms, which means that the expressions under the logarithms are also equal. Formula 4 has been proven.
Example 1.
Calculate 81 log 27 5 log 5 4 .
Solution.
81 = 3 4 , 27 = 3 3 .
log 27 5 = 1/3 log 3 5, log 5 4 = log 3 4 / log 3 5. Therefore,
log 27 5 log 5 4 = 1/3 log 3 5 (log 3 4 / log 3 5) = 1/3 log 3 4.
Then 81 log 27 5 log 5 4 = (3 4) 1/3 log 3 4 = (3 log 3 4) 4/3 = (4) 4/3 = 4 3 √4.
You can complete the following task yourself.
Calculate (8 log 2 3 + 3 1/ log 2 3) - log 0.2 5.
As a hint, 0.2 = 1/5 = 5 -1 ; log 0.2 5 = -1.
Answer: 5.
Example 2.
Calculate (√11) log √3 9- log 121 81 .
Solution.
Let's change the expressions: 9 = 3 2, √3 = 3 1/2, log √3 9 = 4,
121 = 11 2, 81 = 3 4, log 121 81 = 2 log 11 3 (formula 3 was used).
Then (√11) log √3 9- log 121 81 = (11 1/2) 4-2 log 11 3 = (11) 2- log 11 3 = 11 2 / (11) log 11 3 = 11 2 / ( 11 log 11 3) = 121/3.
Example 3.
Calculate log 2 24 / log 96 2 - log 2 192 / log 12 2.
Solution.
We replace the logarithms contained in the example with logarithms with base 2.
log 96 2 = 1/log 2 96 = 1/log 2 (2 5 3) = 1/(log 2 2 5 + log 2 3) = 1/(5 + log 2 3);
log 2 192 = log 2 (2 6 3) = (log 2 2 6 + log 2 3) = (6 + log 2 3);
log 2 24 = log 2 (2 3 3) = (log 2 2 3 + log 2 3) = (3 + log 2 3);
log 12 2 = 1/log 2 12 = 1/log 2 (2 2 3) = 1/(log 2 2 2 + log 2 3) = 1/(2 + log 2 3).
Then log 2 24 / log 96 2 – log 2 192 / log 12 2 = (3 + log 2 3) / (1/(5 + log 2 3)) – ((6 + log 2 3) / (1/( 2 + log 2 3)) =
= (3 + log 2 3) · (5 + log 2 3) – (6 + log 2 3)(2 + log 2 3).
After opening the parentheses and bringing similar terms, we get the number 3. (When simplifying the expression, we can denote log 2 3 by n and simplify the expression
(3 + n) · (5 + n) – (6 + n)(2 + n)).
Answer: 3.
You can complete the following task yourself:
Calculate (log 3 4 + log 4 3 + 2) log 3 16 log 2 144 3.
Here it is necessary to make the transition to base 3 logarithms and factorization of large numbers into prime factors.
Answer:1/2
Example 4.
Given three numbers A = 1/(log 3 0.5), B = 1/(log 0.5 3), C = log 0.5 12 – log 0.5 3. Arrange them in ascending order.
Solution.
Let's transform the numbers A = 1/(log 3 0.5) = log 0.5 3; C = log 0.5 12 – log 0.5 3 = log 0.5 12/3 = log 0.5 4 = -2.
Let's compare them
log 0.5 3 > log 0.5 4 = -2 and log 0.5 3< -1 = log 0,5 2, так как функция у = log 0,5 х – убывающая.
Or 2< log 0,5 3 < -1. Тогда -1 < 1/(log 0,5 3) < -1/2.
Answer. Therefore, the order of placing the numbers is: C; A; IN.
Example 5.
How many integers are in the interval (log 3 1 / 16 ; log 2 6 48).
Solution.
Let us determine between which powers of the number 3 the number 1/16 is located. We get 1/27< 1 / 16 < 1 / 9 .
Since the function y = log 3 x is increasing, then log 3 (1 / 27)< log 3 (1 / 16) < log 3 (1 / 9); -3 < log 3 (1 / 16) < -2.
log 6 48 = log 6 (36 4 / 3) = log 6 36 + log 6 (4 / 3) = 2 + log 6 (4 / 3). Let's compare log 6 (4 / 3) and 1 / 5. And for this we compare the numbers 4/3 and 6 1/5. Let's raise both numbers to the 5th power. We get (4 / 3) 5 = 1024 / 243 = 4 52 / 243< 6. Следовательно,
log 6 (4 / 3)< 1 / 5 . 2 < log 6 48 < 2 1 / 5 . Числа, входящие в двойное неравенство, положительные. Их можно возводить в квадрат. Знаки неравенства при этом не изменятся. Тогда 4 < log 6 2 48 < 4 21 / 25.
Therefore, the interval (log 3 1 / 16 ; log 6 48) includes the interval [-2; 4] and the integers -2 are placed on it; -1; 0; 1; 2; 3; 4.
Answer: 7 integers.
Example 6.
Calculate 3 lglg 2/ lg 3 - lg20.
Solution.
3 lg lg 2/ lg 3 = (3 1/ lg3) lg lg 2 = (3 lо g 3 10) lg lg 2 = 10 lg lg 2 = lg2.
Then 3 lglg2/lg3 - lg 20 = lg 2 – lg 20 = lg 0.1 = -1.
Answer: -1.
Example 7.
It is known that log 2 (√3 + 1) + log 2 (√6 – 2) = A. Find log 2 (√3 –1) + log 2 (√6 + 2).
Solution.
Numbers (√3 + 1) and (√3 – 1); (√6 – 2) and (√6 + 2) are conjugate.
Let us carry out the following transformation of expressions
√3 – 1 = (√3 – 1) · (√3 + 1)) / (√3 + 1) = 2/(√3 + 1);
√6 + 2 = (√6 + 2) · (√6 – 2)) / (√6 – 2) = 2/(√6 – 2).
Then log 2 (√3 – 1) + log 2 (√6 + 2) = log 2 (2/(√3 + 1)) + log 2 (2/(√6 – 2)) =
Log 2 2 – log 2 (√3 + 1) + log 2 2 – log 2 (√6 – 2) = 1 – log 2 (√3 + 1) + 1 – log 2 (√6 – 2) =
2 – log 2 (√3 + 1) – log 2 (√6 – 2) = 2 – A.
Answer: 2 – A.
Example 8.
Simplify and find the approximate value of the expression (log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9.
Solution.
We reduce all logarithms to common ground 10.
(log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9 = (lg 2 / lg 3) (lg 3 / lg 4) (lg 4 / lg 5) (lg 5 / lg 6) · … · (lg 8 / lg 9) · lg 9 = lg 2 ≈ 0.3010 (The approximate value of lg 2 can be found using a table, slide rule or calculator).
Answer: 0.3010.
Example 9.
Calculate log a 2 b 3 √(a 11 b -3) if log √ a b 3 = 1. (In this example, a 2 b 3 is the base of the logarithm).
Solution.
If log √ a b 3 = 1, then 3/(0.5 log a b = 1. And log a b = 1/6.
Then log a 2 b 3√(a 11 b -3) = 1/2 log a 2 b 3 (a 11 b -3) = log a (a 11 b -3) / (2log a (a 2 b 3) ) = (log a a 11 + log a b -3) / (2(log a a 2 + log a b 3)) = (11 – 3log a b) / (2(2 + 3log a b)) Considering that that log a b = 1/6 we get (11 – 3 1 / 6) / (2(2 + 3 1 / 6)) = 10.5/5 = 2.1.
Answer: 2.1.
You can complete the following task yourself:
Calculate log √3 6 √2.1 if log 0.7 27 = a.
Answer: (3 + a) / (3a).
Example 10.
Calculate 6.5 4/ log 3 169 · 3 1/ log 4 13 + log125.
Solution.
6.5 4/ log 3 169 · 3 1/ log 4 13 + log 125 = (13/2) 4/2 log 3 13 · 3 2/ log 2 13 + 2log 5 5 3 = (13/2) 2 log 13 3 3 2 log 13 2 + 6 = (13 log 13 3 / 2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3/2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3 2 /(2 log 13 3) 2) · (2 log 13 3) 2 + 6.
(2 log 13 3 = 3 log 13 2 (formula 4))
We get 9 + 6 = 15.
Answer: 15.
Still have questions? Not sure how to find the value of a logarithmic expression?
To get help from a tutor -.
The first lesson is free!
blog.site, when copying material in full or in part, a link to the original source is required.
As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.
Definition in mathematics
A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number (that is, any positive) “b” to its base “a” is considered to be the power “c” to which the base “a” must be raised in order to ultimately get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.
Types of logarithms
For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three separate types of logarithmic expressions:
- Natural logarithm ln a, where the base is the Euler number (e = 2.7).
- Decimal a, where the base is 10.
- Logarithm of any number b to base a>1.
Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. To obtain the correct values of logarithms, you should remember their properties and the sequence of actions when solving them.
Rules and some restrictions
In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the even root of negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:
- The base “a” must always be greater than zero, and not equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
- if a > 0, then a b >0, it turns out that “c” must also be greater than zero.
How to solve logarithms?
For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.
Now let's represent this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.
To accurately determine the value unknown degree you need to learn how to work with the table of degrees. It looks like this:
As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However, for larger values you will need a power table. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!
Equations and inequalities
It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equality. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.
Given an expression of the following form: log 2 (x-1) > 3 - it is logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.
The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm 2 x = √9) imply one or more specific numerical values in the answer, while when solving an inequality, both the range of acceptable values and the points are determined breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or set of numbers.
Basic theorems about logarithms
When solving primitive tasks of finding the values of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.
- The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
- The logarithm of the product can be represented in the following formula: log d (s 1 *s 2) = log d s 1 + log d s 2. In this case prerequisite is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
- The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
- The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.
This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.
Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;
but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.
Examples of problems and inequalities
The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also a required part of mathematics exams. For admission to university or passing entrance examinations in mathematics you need to know how to solve such problems correctly.
Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, it can be applied to every mathematical inequality or logarithmic equation certain rules. First of all, you should find out whether the expression can be simplified or lead to general appearance. Simplify long ones logarithmic expressions possible if you use their properties correctly. Let's get to know them quickly.
When deciding logarithmic equations, we should determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.
Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. For solutions of natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.
How to Use Logarithm Formulas: With Examples and Solutions
So, let's look at examples of using the basic theorems about logarithms.
- The property of the logarithm of a product can be used in tasks where it is necessary to expand great importance numbers b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
- log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values out of the sign of the logarithm.
Assignments from the Unified State Exam
Logarithms are often found in entrance exams, especially many logarithmic problems in the Unified State Exam ( State exam for all school leavers). Typically, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge of the topic “Natural logarithms”.
Examples and solutions to problems are taken from official Unified State Exam options. Let's see how such tasks are solved.
Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.
- It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
- All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.
When converting expressions with logarithms, the listed equalities are used both from right to left and from left to right.
It is worth noting that it is not necessary to memorize the consequences of the properties: when carrying out transformations, you can get by with the basic properties of logarithms and other facts (for example, the fact that for b≥0), from which the corresponding consequences follow. " By-effect"This approach only manifests itself in the fact that the solution will be a little longer. For example, in order to do without the consequence, which is expressed by the formula 
, and starting only from the basic properties of logarithms, you will have to carry out a chain of transformations of the following form: 
.
The same can be said about the last property from the list above, which is answered by the formula 
, since it also follows from the basic properties of logarithms. The main thing to understand is that it is always possible for the power of a positive number with a logarithm in the exponent to swap the base of the power and the number under the logarithm sign. To be fair, we note that examples implying the implementation of transformations of this kind are rare in practice. We will give a few examples below in the text.
Converting numeric expressions with logarithms
We have remembered the properties of logarithms, now it’s time to learn how to apply them in practice to transform expressions. It is natural to start with converting numerical expressions rather than expressions with variables, since they are more convenient and easier to learn the basics. That's what we'll do, and we'll start with a very simple examples, to learn how to choose the desired property of the logarithm, but we will gradually complicate the examples, up to the point when, to obtain final result you will need to apply several properties in a row.
Selecting the desired property of logarithms
There are many properties of logarithms, and it is clear that you need to be able to choose the appropriate one from them, which in this particular case will lead to the required result. Usually this is not difficult to do by comparing the type of converted logarithm or expression with the types of left and right parts of formulas expressing the properties of logarithms. If left or right part one of the formulas coincides with a given logarithm or expression, then, most likely, it is this property that should be used during the transformation. The following examples this is clearly demonstrated.
Let's start with examples of transforming expressions using the definition of a logarithm, which corresponds to the formula a log a b =b, a>0, a≠1, b>0.
Example.
Calculate, if possible: a) 5 log 5 4, b) 10 log(1+2·π), c) 
, d) 2 log 2 (−7) , e) .
Solution.
In the example under the letter a) the structure a log a b is clearly visible, where a=5, b=4. These numbers satisfy the conditions a>0, a≠1, b>0, so you can safely use the equality a log a b =b. We have 5 log 5 4=4 .
b) Here a=10, b=1+2·π, the conditions a>0, a≠1, b>0 are met. In this case, the equality 10 log(1+2·π) =1+2·π takes place.
c) And in this example we are dealing with a degree of the form a log a b, where and b=ln15. So 
.
Despite belonging to the same type a log a b (here a=2, b=−7), the expression under the letter g) cannot be converted using the formula a log a b =b. The reason is that it is meaningless because it contains a negative number under the logarithm sign. Moreover, the number b=−7 does not satisfy the condition b>0, which makes it impossible to resort to the formula a log a b =b, since it requires the fulfillment of the conditions a>0, a≠1, b>0. So, we can't talk about calculating the value of 2 log 2 (−7) . In this case, writing 2 log 2 (−7) =−7 would be an error.
Similarly, in the example under letter e) it is impossible to give a solution of the form 
, since the original expression does not make sense.
Answer:
a) 5 log 5 4 =4, b) 10 log(1+2·π) =1+2·π, c) , d), e) expressions do not make sense.
It is often useful to transform in which positive number is represented as a power of some positive and non-unit number with a logarithm in the exponent. It is based on the same definition of the logarithm a log a b =b, a>0, a≠1, b>0, but the formula is applied from right to left, that is, in the form b=a log a b. For example, 3=e ln3 or 5=5 log 5 5 .
Let's move on to using the properties of logarithms to transform expressions.
Example.
Find the value of the expression: a) log −2 1, b) log 1 1, c) log 0 1, d) log 7 1, e) ln1, f) log1, g) log 3.75 1, h) log 5 π 7 1 .
Solution.
In the examples under the letters a), b) and c) the expressions log −2 1, log 1 1, log 0 1 are given, which do not make sense, since the base of the logarithm should not contain a negative number, zero or one, because we have defined logarithm only for a base that is positive and different from unity. Therefore, in examples a) - c) there can be no question of finding the meaning of the expression.
In all other tasks, obviously, the bases of the logarithms contain positive and non-unity numbers 7, e, 10, 3.75 and 5·π 7, respectively, and under the signs of the logarithms there are units everywhere. And we know the property of the logarithm of unity: log a 1=0 for any a>0, a≠1. Thus, the values of expressions b) – e) are equal to zero.
Answer:
a), b), c) expressions do not make sense, d) log 7 1=0, e) ln1=0, f) log1=0, g) log 3.75 1=0, h) log 5 e 7 1=0 .
Example.
Calculate: a) , b) lne , c) lg10 , d) log 5 π 3 −2 (5 π 3 −2), e) log −3 (−3) , f) log 1 1 .
Solution.
It is clear that we have to use the property of the logarithm of the base, which corresponds to the formula log a a=1 for a>0, a≠1. Indeed, in the tasks under all the letters, the number under the logarithm sign coincides with its base. Thus, I would like to immediately say that the value of each of the given expressions is 1. However, you should not rush to conclusions: in tasks under the letters a) - d) the values of the expressions are really equal to one, and in tasks e) and f) the original expressions do not make sense, so it cannot be said that the values of these expressions are equal to 1.
Answer:
a) , b) lne=1 , c) lg10=1 , d) log 5 π 3 −2 (5 π 3 −2)=1, e), f) expressions do not make sense.
Example.
Find the value: a) log 3 3 11, b) 
, c) , d) log −10 (−10) 6 .
Solution.
Obviously, under the signs of logarithms there are some powers of the base. Based on this, we understand that here we will need the property of the degree of the base: log a a p =p, where a>0, a≠1 and p is any real number. Taking this into account, we have the following results: a) log 3 3 11 =11, b) 
, V) 
. Is it possible to write a similar equality for the example under the letter d) of the form log −10 (−10) 6 =6? No, you can't, because the expression log −10 (−10) 6 doesn't make sense.
Answer:
a) log 3 3 11 =11, b) , V) , d) the expression does not make sense.
Example.
Present the expression as a sum or difference of logarithms using the same base: a) 
, b) , c) log((−5)·(−12)) .
Solution.
a) Under the sign of the logarithm there is a product, and we know the property of the logarithm of the product log a (x·y)=log a x+log a y, a>0, a≠1, x>0, y>0. In our case, the number in the base of the logarithm and the numbers in the product are positive, that is, they satisfy the conditions of the selected property, therefore, we can safely apply it: 
.
b) Here we use the property of the quotient logarithm, where a>0, a≠1, x>0, y>0. In our case, the base of the logarithm is a positive number e, the numerator and denominator π are positive, which means they satisfy the conditions of the property, so we have the right to use the chosen formula: 
.
c) First, note that the expression log((−5)·(−12)) makes sense. But at the same time, for it we do not have the right to apply the formula for the logarithm of the product log a (x y)=log a x+log a y, a>0, a≠1, x>0, y>0, since the numbers are −5 and −12 – negative and do not satisfy the conditions x>0, y>0. That is, you cannot carry out such a transformation: log((−5)·(−12))=log(−5)+log(−12). So what should we do? In such cases, the original expression needs a preliminary transformation to avoid negative numbers. About similar cases of transforming expressions with negative numbers under the sign of the logarithm, we will talk in detail in one of them, but for now we will give a solution to this example, which is clear in advance and without explanation: log((−5)·(−12))=log(5·12)=log5+lg12.
Answer:
A) , b) , c) log((−5)·(−12))=log5+lg12.
Example.
Simplify the expression: a) log 3 0.25+log 3 16+log 3 0.5, b) .
Solution.
Here we will be helped by all the same properties of the logarithm of the product and the logarithm of the quotient that we used in previous examples, only now we will apply them from right to left. That is, we transform the sum of logarithms into the logarithm of the product, and the difference of logarithms into the logarithm of the quotient. We have
A) log 3 0.25+log 3 16+log 3 0.5=log 3 (0.25 16 0.5)=log 3 2.
b) 
.
Answer:
A) log 3 0.25+log 3 16+log 3 0.5=log 3 2, b) 
.
Example.
Get rid of the degree under the logarithm sign: a) log 0.7 5 11, b) 
, c) log 3 (−5) 6 .
Solution.
It is easy to see that we are dealing with expressions of the form log a b p . The corresponding property of the logarithm has the form log a b p =p·log a b, where a>0, a≠1, b>0, p - any real number. That is, if the conditions a>0, a≠1, b>0 are met, from the logarithm of the power log a b p we can proceed to the product p·log a b. Let's carry out this transformation with the given expressions.
a) In this case a=0.7, b=5 and p=11. So log 0.7 5 11 =11·log 0.7 5.
b) Here, the conditions a>0, a≠1, b>0 are satisfied. That's why 
c) The expression log 3 (−5) 6 has the same structure log a b p , a=3 , b=−5 , p=6 . But for b the condition b>0 is not satisfied, which makes it impossible to use the formula log a b p =p·log a b . So what, you can’t cope with the task? It is possible, but a preliminary transformation of the expression is required, which we will discuss in detail below in the paragraph under the heading. The solution will be like this: log 3 (−5) 6 =log 3 5 6 =6 log 3 5.
Answer:
a) log 0.7 5 11 =11 log 0.7 5 ,
b)
c) log 3 (−5) 6 =6·log 3 5.
Quite often, when carrying out transformations, the formula for the logarithm of a power has to be applied from right to left in the form p·log a b=log a b p (the same conditions must be met for a, b and p). For example, 3·ln5=ln5 3 and log2·log 2 3=log 2 3 lg2.
Example.
a) Calculate the value of log 2 5 if it is known that log2≈0.3010 and log5≈0.6990. b) Express the fraction as a logarithm to base 3.
Solution.
a) The formula for transition to a new logarithm base allows us to present this logarithm as a ratio of decimal logarithms, the values of which are known to us: . All that remains is to carry out the calculations, we have 
.
b) Here it is enough to use the formula for moving to a new base, and apply it from right to left, that is, in the form 
. We get 
.
Answer:
a) log 2 5≈2.3223, b) .
At this stage, we have quite thoroughly examined the transformation of the simplest expressions using the basic properties of logarithms and the definition of a logarithm. In these examples, we had to apply one property and nothing more. Now, with a clear conscience, you can move on to examples, the transformation of which requires the use of several properties of logarithms and other additional transformations. We will deal with them in the next paragraph. But before that, let us briefly look at examples of the application of consequences from the basic properties of logarithms.
Example.
a) Get rid of the root under the logarithm sign. b) Convert the fraction to base 5 logarithm. c) Free yourself from powers under the sign of the logarithm and in its base. d) Calculate the value of the expression 
. e) Replace the expression with a power with base 3.
Solution.
a) If we recall the corollary from the property of the logarithm of the degree 
, then you can immediately give the answer: 
.
b) Here we use the formula 
from right to left, we have 
.
c) B in this case the result is given by the formula . We get 
.
d) And here it is enough to apply the corollary to which the formula corresponds 
. So 
.
e) Property of the logarithm allows us to achieve desired result: 
.
Answer:
A) . b) . V) 
. G) 
. d) .
Consecutive application of several properties
Real tasks on transforming expressions using the properties of logarithms are usually more complicated than those we dealt with in the previous paragraph. In them, as a rule, the result is not obtained in one step, but the solution already consists in the sequential application of one property after another, together with additional identical transformations, such as opening parentheses, bringing similar terms, reducing fractions, etc. So let's get closer to such examples. There is nothing complicated about this, the main thing is to act carefully and consistently, observing the order of actions.
Example.
Calculate the value of an expression (log 3 15−log 3 5) 7 log 7 5.
Solution.
The difference between the logarithms in parentheses, according to the property of the quotient logarithm, can be replaced by the logarithm log 3 (15:5), and then calculate its value log 3 (15:5)=log 3 3=1. And the value of the expression 7 log 7 5 by definition of a logarithm is equal to 5. Substituting these results into the original expression, we get (log 3 15−log 3 5) 7 log 7 5 =1 5=5.
Here is a solution without explanation:
(log 3 15−log 3 5) 7 log 7 5 =log 3 (15:5) 5=
=log 3 3·5=1·5=5 .
Answer:
(log 3 15−log 3 5) 7 log 7 5 =5.
Example.
What is the value of the numerical expression log 3 log 2 2 3 −1?
Solution.
We first transform the logarithm under the logarithm sign using the formula for the logarithm of the power: log 2 2 3 =3. Thus, log 3 log 2 2 3 =log 3 3 and then log 3 3=1. So log 3 log 2 2 3 −1=1−1=0 .
Answer:
log 3 log 2 2 3 −1=0 .
Example.
Simplify the expression.
Solution.
The formula for moving to a new logarithm base allows the ratio of logarithms to one base to be represented as log 3 5. In this case, the original expression will take the form . By definition of the logarithm 3 log 3 5 =5, that is 
, and the value of the resulting expression, by virtue of the same definition of the logarithm, is equal to two.
Here is a short version of the solution that is usually given: 
.
Answer:
.
To smoothly transition to the information in the next paragraph, let's take a look at the expressions 5 2+log 5 3, and log0.01. Their structure does not fit any of the properties of logarithms. So what happens, they cannot be converted using the properties of logarithms? It is possible if you carry out preliminary transformations that prepare these expressions for the application of the properties of logarithms. So 5 2+log 5 3 =5 2 5 log 5 3 =25 3=75, 
and log0.01=log10 −2 =−2. Next we will look in detail at how such expression preparation is carried out.
Preparing Expressions to Use the Properties of Logarithms
Logarithms in the expression being converted very often differ in the structure of the notation from the left and right parts of formulas that correspond to the properties of logarithms. But no less often, the transformation of these expressions involves the use of the properties of logarithms: their use only requires preliminary preparation. And this preparation consists of carrying out certain identity transformations, bringing logarithms to a form convenient for applying the properties.
To be fair, we note that almost any transformation of expressions can act as preliminary transformations, from the banal reduction of similar terms to the application trigonometric formulas. This is understandable, since the expressions being converted can contain any mathematical objects: brackets, modules, fractions, roots, powers, etc. Thus, one must be prepared to perform any necessary transformation in order to further be able to take advantage of the properties of logarithms.
Let us say right away that at this point we do not set ourselves the task of classifying and analyzing all conceivable preliminary transformations that would allow us to subsequently apply the properties of logarithms or the definition of a logarithm. Here we will focus on only four of them, which are the most typical and most often encountered in practice.
And now about each of them in detail, after which, within the framework of our topic, all that remains is to understand the transformation of expressions with variables under the signs of logarithms.
Identification of powers under the logarithm sign and at its base
Let's start right away with an example. Let us have a logarithm. Obviously, in this form its structure is not conducive to the use of the properties of logarithms. Is it possible to somehow transform this expression to simplify it, and even better to calculate its value? To answer this question, let's take a closer look at the numbers 81 and 1/9 in the context of our example. Here it is easy to notice that these numbers can be represented as a power of 3, indeed, 81 = 3 4 and 1/9 = 3 −2. In this case, the original logarithm is presented in the form and it becomes possible to apply the formula . So, 
.
Analysis of the analyzed example gives rise to the following thought: if possible, you can try to isolate the degree under the sign of the logarithm and in its base in order to apply the property of the logarithm of the degree or its consequences. It remains only to figure out how to distinguish these degrees. Let's give some recommendations on this issue.
Sometimes it is quite obvious that the number under the logarithm sign and/or in its base represents some integer power, as in the example discussed above. Almost constantly we have to deal with powers of two, which are well familiar: 4=2 2, 8=2 3, 16=2 4, 32=2 5, 64=2 6, 128=2 7, 256=2 8, 512= 2 9, 1024=2 10. The same can be said about the powers of three: 9 = 3 2, 27 = 3 3, 81 = 3 4, 243 = 3 5, ... In general, it won’t hurt if you have before your eyes table of degrees natural numbers within a dozen. It is also not difficult to work with integer powers of ten, one hundred, thousand, etc.
Example.
Calculate the value or simplify the expression: a) log 6 216, b) , c) log 0.000001 0.001.
Solution.
a) Obviously, 216=6 3, so log 6 216=log 6 6 3 =3.
b) The table of powers of natural numbers allows you to represent the numbers 343 and 1/243 as powers 7 3 and 3 −4, respectively. Therefore, the following transformation of a given logarithm is possible: 
c) Since 0.000001=10 −6 and 0.001=10 −3, then log 0.000001 0.001=log 10 −6 10 −3 =(−3)/(−6)=1/2.
Answer:
a) log 6 216=3, b) 
, c) log 0.000001 0.001=1/2.
In more difficult cases to distinguish powers of numbers one has to resort to .
Example.
Convert the expression to more simple view log 3 648 log 2 3 .
Solution.
Let's look at what the factorization of 648 is:
That is, 648=2 3 ·3 4. Thus, log 3 648 log 2 3=log 3 (2 3 3 4) log 2 3.
Now we convert the logarithm of the product into the sum of logarithms, after which we apply the properties of the logarithm of the power:
log 3 (2 3 3 4)log 2 3=(log 3 2 3 +log 3 3 4)log 2 3=
=(3·log 3 2+4)·log 2 3 .
By virtue of a corollary from the property of the logarithm of the power, which corresponds to the formula , the product log32·log23 is the product of , and, as is known, it is equal to one. Taking this into account, we get 3 log 3 2 log 2 3+4 log 2 3=3 1+4 log 2 3=3+4 log 2 3.
Answer:
log 3 648 log 2 3=3+4 log 2 3.
Quite often, expressions under the sign of the logarithm and in its base represent products or ratios of the roots and/or powers of some numbers, for example, , . Such expressions can be expressed as powers. To do this, a transition is made from roots to powers, and and are used. These transformations make it possible to isolate the powers under the sign of the logarithm and in its base, and then apply the properties of logarithms.
Example.
Calculate: a) 
, b) .
Solution.
a) The expression in the base of the logarithm is the product of powers with on the same grounds, by the corresponding property of powers we have 5 2 ·5 −0.5 ·5 −1 =5 2−0.5−1 =5 0.5.
Now let’s transform the fraction under the logarithm sign: we’ll move from the root to the power, after which we’ll use the property of the ratio of powers with the same bases: 
.
It remains to substitute the obtained results into the original expression, use the formula and finish the transformation: 
b) Since 729 = 3 6 and 1/9 = 3 −2, the original expression can be rewritten as .
Next, we apply the property of the root of a power, move from the root to the power, and use the property of the ratio of powers to convert the base of the logarithm to a power: 
.
Considering last result, we have 
.
Answer:
A) 
, b) .
It is clear that in general case To obtain powers under the sign of the logarithm and in its base, various transformations of various expressions may be required. Let's give a couple of examples.
Example.
What is the meaning of the expression: a) 
, b) 
.
Solution.
We further note that the given expression has the form log A B p , where A=2, B=x+1 and p=4. We transformed numerical expressions of this type according to the property of the logarithm of the power log a b p =p·log a b , therefore, with a given expression I want to do the same, and move from log 2 (x+1) 4 to 4·log 2 (x+1) . Now let's calculate the value of the original expression and the expression obtained after the transformation, for example, when x=−2. We have log 2 (−2+1) 4 =log 2 1=0 , and 4 log 2 (−2+1)=4 log 2 (−1)- a meaningless expression. This raises a logical question: “What did we do wrong?”
And the reason is this: we performed the transformation log 2 (x+1) 4 =4·log 2 (x+1) , based on the formula log a b p =p·log a b , but we have the right to apply this formula only if the conditions a >0, a≠1, b>0, p - any real number. That is, the transformation we have done takes place if x+1>0, which is the same as x>−1 (for A and p, the conditions are met). However, in our case, the ODZ of variable x for the original expression consists not only of the interval x>−1, but also of the interval x<−1 . Но для x<−1 мы не имели права осуществлять преобразование по выбранной формуле.
The need to take into account DL
Let's continue to analyze the transformation of the expression we have chosen log 2 (x+1) 4 , and now let's see what happens to the ODZ when moving to the expression 4 · log 2 (x+1) . In the previous paragraph, we found the ODZ of the original expression - this is the set (−∞, −1)∪(−1, +∞) . Now let’s find the range of acceptable values of the variable x for the expression 4·log 2 (x+1) . It is determined by the condition x+1>0, which corresponds to the set (−1, +∞). It is obvious that when moving from log 2 (x+1) 4 to 4·log 2 (x+1), the range of permissible values narrows. And we agreed to avoid transformations that lead to a narrowing of DL, as this can lead to various negative consequences.
Here it is worth noting that it is useful to control the OA at each step of the transformation and prevent its narrowing. And if suddenly at some stage of the transformation there was a narrowing of the DL, then it is worth looking very carefully at whether this transformation is permissible and whether we had the right to carry out it.
To be fair, let’s say that in practice we usually have to work with expressions in which the variable value of variables is such that, when carrying out transformations, we can use the properties of logarithms without restrictions in the form already known to us, both from left to right and from right to left. You quickly get used to this, and you begin to carry out transformations mechanically, without thinking about whether it was possible to carry them out. And at such moments, as luck would have it, more complex examples slip through in which careless application of the properties of logarithms leads to errors. So you need to always be on the lookout and make sure that there is no narrowing of the ODZ.
It would not hurt to separately highlight the main transformations based on the properties of logarithms, which must be carried out very carefully, which can lead to a narrowing of the OD, and as a result – to errors:
Some transformations of expressions based on the properties of logarithms can also lead to the opposite - expansion of the ODZ. For example, the transition from 4·log 2 (x+1) to log 2 (x+1) 4 expands the ODZ from the set (−1, +∞) to (−∞, −1)∪(−1, +∞) . Such transformations take place if we remain within the framework of the ODZ for the original expression. So the just mentioned transformation 4·log 2 (x+1)=log 2 (x+1) 4 takes place on the ODZ of the variable x for the original expression 4·log 2 (x+1), that is, for x+1> 0, which is the same as (−1, +∞).
Now that we have discussed the nuances that you need to pay attention to when transforming expressions with variables using the properties of logarithms, it remains to figure out how to correctly carry out these transformations.
X+2>0 . Does it work in our case? To answer this question, let's take a look at the ODZ of the variable x. It is determined by the system of inequalities 
, which is equivalent to the condition x+2>0 (if necessary, see the article solving systems of inequalities). Thus, we can safely apply the property of the logarithm of the power.
We have
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =
=3·7·log(x+2)−log(x+2)−5·4·log(x+2)=
=21 log(x+2)−log(x+2)−20 log(x+2)=
=(21−1−20)·log(x+2)=0 .
You can act differently, since the ODZ allows you to do this, for example like this: 
Answer:
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =0.
But what to do when the conditions accompanying the properties of logarithms are not met in the ODZ? We will understand this with examples.
Let us be required to simplify the expression log(x+2) 4 − log(x+2) 2 . The transformation of this expression, unlike the expression from the previous example, does not allow free use of the property of the logarithm of the power. Why? The ODZ of variable x in this case is the union of two intervals x>−2 and x<−2 . При x>−2 we can easily apply the property of the logarithm of a power and act as in the example above: log(x+2) 4 −log(x+2) 2 =4 log(x+2)−2 log(x+2)=2 log(x+2). But the ODZ contains one more interval x+2<0 , для которого последнее преобразование будет некорректно. Что же делать при x+2<0 ? В подобных случаях на помощь приходит . Определение модуля позволяет выражение x+2 при x+2<0 представить как −|x+2| . Тогда при x+2<0 от lg(x+2) 4 −lg(x+2) 2 переходим к log(−|x+2|) 4 −log(−|x+2|) 2 and further due to the properties of the degree k lg|x+2| 4 −lg|x+2| 2. The resulting expression can be transformed using the property of the logarithm of a power, since |x+2|>0 for any value of the variable. We have log|x+2| 4 −lg|x+2| 2 =4·lg|x+2|−2·lg|x+2|=2·lg|x+2|. Now you can free yourself from the module, since it has done its job. Since we carry out the transformation at x+2<0 , то 2·lg|x+2|=2·lg(−(x+2)) . Итак, можно считать, что мы справились с поставленной задачей. Ответ: . Полученный результат можно записать компактно с использованием модуля как .
Let's look at one more example so that working with modules becomes familiar. Let us conceive from the expression 
go to the sum and difference of logarithms of linear binomials x−1, x−2 and x−3. First we find the ODZ: 
On the interval (3, +∞) the values of the expressions x−1, x−2 and x−3 are positive, so we can easily apply the properties of the logarithm of the sum and difference: 
And on the interval (1, 2) the values of the expression x−1 are positive, and the values of the expressions x−2 and x−3 are negative. Therefore, on the considered interval we represent x−2 and x−3 using the modulus as −|x−2| and −|x−3| respectively. Wherein 
Now we can apply the properties of the logarithm of the product and the quotient, since on the considered interval (1, 2) the values of the expressions x−1 , |x−2| and |x−3| - positive.
We have 
The results obtained can be combined:
In general, similar reasoning allows, based on the formulas for the logarithm of the product, ratio and degree, to obtain three practically useful results, which are quite convenient to use:
- The logarithm of the product of two arbitrary expressions X and Y of the form log a (X·Y) can be replaced by the sum of logarithms log a |X|+log a |Y| , a>0 , a≠1 .
- The logarithm of a particular form log a (X:Y) can be replaced by the difference of logarithms log a |X|−log a |Y| , a>0, a≠1, X and Y are arbitrary expressions.
- From the logarithm of some expression B to an even power p of the form log a B p we can go to the expression p·log a |B| , where a>0, a≠1, p is an even number and B is an arbitrary expression.
Similar results are given, for example, in the instructions for solving exponential and logarithmic equations in the collection of problems in mathematics for those entering universities, edited by M. I. Skanavi.
Example.
Simplify the expression 
.
Solution.
It would be good to apply the properties of the logarithm of the power, sum and difference. But can we do this here? To answer this question we need to know the DZ.
Let's define it: 
It is quite obvious that the expressions x+4, x−2 and (x+4) 13 in the range of permissible values of the variable x can take on both positive and negative values. Therefore, we will have to act through modules. 
The module properties allow you to rewrite it as , so 
Also, nothing prevents you from using the property of the logarithm of a power, and then bringing similar terms: 
Another sequence of transformations leads to the same result: 
and since on the ODZ the expression x−2 can take both positive and negative values, then when taking an even exponent 14
